Answer:
Area of the gray region = 64 square units
Step-by-step explanation:
Area of the white square = 64 square units
Length of a side of the given square = ![\sqrt{\text{Area}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Ctext%7BArea%7D%7D)
= ![\sqrt{64}](https://tex.z-dn.net/?f=%5Csqrt%7B64%7D)
= 8 units
By Pythagoras theorem,
Length of diagonal DB = ![\sqrt{BC^2+CD^2}](https://tex.z-dn.net/?f=%5Csqrt%7BBC%5E2%2BCD%5E2%7D)
DB = ![\sqrt{8^2+8^2}](https://tex.z-dn.net/?f=%5Csqrt%7B8%5E2%2B8%5E2%7D)
= ![8\sqrt{2}](https://tex.z-dn.net/?f=8%5Csqrt%7B2%7D)
AD = DB =
[Given]
OD = OC = ![\frac{1}{2}(DB)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28DB%29)
= 4√2
Therefore, AO = AD + OD = 8√2 + 4√2
= 12√2
Area of ΔACD = Area of ΔAOC - Area of ΔCOD
= ![\frac{1}{2}(AO)(OC)-\frac{1}{2}(CO)(OD)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28AO%29%28OC%29-%5Cfrac%7B1%7D%7B2%7D%28CO%29%28OD%29)
= ![\frac{1}{2}[(12\sqrt{2}\times 4\sqrt{2})-(4\sqrt{2})^2]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B%2812%5Csqrt%7B2%7D%5Ctimes%204%5Csqrt%7B2%7D%29-%284%5Csqrt%7B2%7D%29%5E2%5D)
= ![\frac{1}{2}[96-32]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B96-32%5D)
= 32
Therefore, area of gray part = Area of ΔACD + Area of ΔAED
= 32 + 32
= 64 square units
I think that the answer is B,
Answer:
I think c is the answers but I could be wrong
Answer:
A. 156 students are in choir
Step-by-step explanation:
It states that for every 3 students in choir equals 1 student in band which give you the equation 1x3. So if band has 52 students then your new equation is 52x3 which is 156! Hope this helps!:)